Question

If $a$ and $b$ are positive numbers such that $a>b,$ then the minimum value of $a\sec θ−b\tan θ\left(0<θ<\frac{π}{2}\right)$ is

• A. $\frac{1}{\sqrt{a^2−b^2}}$
• B. $\frac{1}{\sqrt{a^2+b^2}}$
• C. $\sqrt{a^2+b^2}$
• D. $\sqrt{a^2−b^2}$
• E. $a^2−b^2$

Answer 1

Answer: (D)

Let $y=a\sec θ−b\tan θ$
$⇒$ $\frac{dy}{dθ}=a\sec θ\tan θ−b{\sec }^2θ$
Put $\frac{dy}{dθ}=0⇒\sec θ(a\tan θ−b\sec θ)=0$
$⇒$ $\sin θ=\frac{b}{a}$ $(âˆ\mu \sec θî€=0)$ Now, $\frac{d^{2}y}{d{θ}^2}>0,$ at $\sin θ=\frac{b}{a}$
$∴$ Minimum value is
$y=a.\frac{a}{\sqrt{a^2−b^2}}−b.\frac{b}{\sqrt{a^2−b^2}}$
$=\sqrt{a^2−b^2}$